10

我想实例化一个可变参数模板类，该类对包中的每种类型Store<TArgs...>都有一个。std::vectorTArgs...

template<typename... TArgs> class Store {
    // obviously not valid code
    // assuming each type of TArgs... has a `unsigned int` id that can be
    // retrieved with getId<T>()
    std::array<sizeof...(TArgs), std::vector<TArgs...>> bags;

    template<typename T> void add(T mValue) { 
        bags[getId<T>()].push_back(mValue); 
    }

    template<typename T> std::vector<T>& get() { 
        return bags[getId<T>()]; 
    }
};

假设我有一个Store<int, float, double>. 我显然知道，在编译时，它将能够存储int,float和double值。

我可以使用模板专业化：

template<> class Store<int, float, double> {
    std::vector<int> vi;
    std::vector<float> vf;
    std::vector<double> vd;

    template<typename T> void add(T);
    template<> void add<int>(int mValue) { vi.push_back(mValue); }
    template<> void add<float>(float mValue) { vf.push_back(mValue); }
    template<> void add<double>(double mValue) { vd.push_back(mValue); }
    // ...
};

...但这需要手写所有可能的类型组合，并且不适用于用户定义的类型。

我相信编译器知道生成类所需的一切，比如Store<int, float, double>使用可变参数模板——有没有办法真正表达这个意图？

4

2 回答 2

10

以下应该做你想要的：

#include <type_traits>
#include <vector>
#include <tuple>
#include <iostream>

// indices are a classic
template< std::size_t... Ns >
struct indices
{
    using next = indices< Ns..., sizeof...( Ns ) >;
};

template< std::size_t N >
struct make_indices
{
    using type = typename make_indices< N - 1 >::type::next;
};

template<>
struct make_indices< 0 >
{
    using type = indices<>;
};

// we need something to find a type's index within a list of types
template<typename T, typename U, std::size_t=0>
struct index {};

template<typename T, typename... Us, std::size_t N>
struct index<T,std::tuple<T,Us...>,N>
: std::integral_constant<std::size_t, N> {};

template<typename T, typename U, typename... Us, std::size_t N>
struct index<T,std::tuple<U,Us...>,N>
: index<T,std::tuple<Us...>,N+1> {};

// we need a way to remove duplicate types from a list of types
template<typename T,typename I=void> struct unique;

// step 1: generate indices
template<typename... Ts>
struct unique< std::tuple<Ts...>, void >
: unique< std::tuple<Ts...>, typename make_indices<sizeof...(Ts)>::type >
{
};

// step 2: remove duplicates. Note: No recursion here!
template<typename... Ts, std::size_t... Is>
struct unique< std::tuple<Ts...>, indices<Is...> >
{
    using type = decltype( std::tuple_cat( std::declval<
        typename std::conditional<index<Ts,std::tuple<Ts...>>::value==Is,std::tuple<Ts>,std::tuple<>>::type
>()... ) );
};

// a helper to turn Ts... into std::vector<Ts>...
template<typename> struct vectorize;

template<typename... Ts>
struct vectorize<std::tuple<Ts...>>
{
    using type = std::tuple< std::vector<Ts>... >;
};

// now you can easily use it to define your Store
template<typename... Ts> class Store
{
    using Storage = typename vectorize<typename unique<std::tuple<Ts...>>::type>::type;
    Storage storage;

    template<typename T>
    decltype(std::get<index<T,typename unique<std::tuple<Ts...>>::type>::value>(storage))
    slot()
    {
        return std::get<index<T,typename unique<std::tuple<Ts...>>::type>::value>(storage);
    }

public:
    template<typename T> void add(T mValue) { 
        slot<T>().push_back(mValue); 
    }

    template<typename T> std::vector<T>& get() { 
        return slot<T>();
    }    
};

int main()
{
    Store<int,int,double,int,double> store;
    store.add(42);
    store.add(3.1415);
    store.add(21);
    std::cout << store.get<int>().size() << std::endl; 
    std::cout << store.get<double>().size() << std::endl; 
}

现场示例（没有评论）

于 2013-10-19T08:42:08.770 回答
6

在 C++14 中，std::tuple如果没有多个相同类型的元素，则可以按类型访问 的元素。因此，您应该可以这样写：

template<typename... TArgs>
struct Store {

    std::tuple<std::vector<TArgs>...> bags;

    template<typename T>
    void add(T mValue) { 
        get<T>().push_back(mValue); 
    }

    template<typename T>
    std::vector<T>& get() { 
        return std::get<std::vector<T>>(bags); 
    }
};
于 2013-10-19T08:51:29.753 回答